GATE 2024 EC Q.44: solution and answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Read the two figures. $x(t)$ is zero outside $[-1,0]$ and forms a triangle of height $-1$ with base 1. The tick under its minimum is printed as '0.5', but the tick sits halfway between $-1$ and $0$, so the minimum is at $t=-\frac{1}{2}$. That is the only reading that fits a triangle on $[-1,0]$. $y(t)$ is a triangle of height $+1$ with base 4, running from $t=-2$ to $t=2$ and centred at $t=0$. So $y$ is $x$ with its sign flipped, stretched 4 times in time, and moved so that its centre sits at the origin.

2. Try $y(t)=-x\left(\frac{t}{4}-\frac{1}{2}\right)$. The argument $\frac{t}{4}-\frac{1}{2}$ is a linear function of $t$, so it sends straight segments to straight segments and corners to corners. The two sides are therefore equal if they agree at the three corners and are both zero outside. At $t=0$: $-x(-\frac{1}{2})=-(-1)=1$. At $t=2$: $-x(0)=0$. At $t=-2$: $-x(-1)=0$. For $|t|>2$ the argument falls outside $[-1,0]$, so the right side is zero, just like $y(t)$. The relation holds.

3. Rewrite the argument as $\frac{t}{4}-\frac{1}{2}=\frac{t-2}{4}$. Then $y(t)=-g(t-2)$ with $g(t)=x(t/4)$.

4. Time-scaling property: $x(at)\leftrightarrow \frac{1}{|a|}X\left(\frac{f}{a}\right)$. With $a=\frac{1}{4}$, $G(f)=4X(4f)$.

5. Time-shift property: $g(t-t_0)\leftrightarrow G(f)e^{-j2\pi f{t}_0}$. With $t_0=2$ and the minus sign carried along: $$Y(f)=-4X(4f)e^{-j2\pi f(2)}=-4X(4f)e^{-j4\pi f}$$

6. Check by doing the operations in the other order. First shift: $x(t-\frac{1}{2})\leftrightarrow X(f)e^{-j\pi f}$. Then replace $t$ by $t/4$. Every $f$ becomes $4f$ and the whole transform is multiplied by 4, giving $4X(4f)e^{-j\pi (4f)}=4X(4f)e^{-j4\pi f}$. With the minus sign this is the same result as before.

7. Check with closed forms. Let $\Lambda (t)=1-|t|$ for $|t|\le 1$ and $0$ otherwise, so that $\Lambda (t)\leftrightarrow {\text{sinc}}^2(f)$ with $\text{sinc}(u)=\frac{\sin \pi u}{\pi u}$. Then $x(t)=-\Lambda (2(t+\frac{1}{2}))$, so $X(f)=-\frac{1}{2}{\text{sinc}}^2(f/2)e^{j\pi f}$. Substituting: $$-4X(4f)e^{-j4\pi f}=2{\text{sinc}}^2(2f)e^{j4\pi f}{e}^{-j4\pi f}=2{\text{sinc}}^2(2f)$$ Computed directly, $y(t)=\Lambda (t/2)\leftrightarrow 2{\text{sinc}}^2(2f)$. The two agree. Option (A) would instead give $2{\text{sinc}}^2(2f)e^{j3\pi f}$, which is not real, but $y(t)$ is real and even, so its transform must be real. The answer is (B).